3.1091 \(\int \frac {1}{x^2 (-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=149 \[ -\frac {\sqrt [4]{3 x^2-1}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{2 x} \]

[Out]

-1/2*(3*x^2-1)^(1/4)/x+1/8*arctan(1/2*x*6^(1/2)/(3*x^2-1)^(1/4))*6^(1/2)-1/8*arctanh(1/2*x*6^(1/2)/(3*x^2-1)^(
1/4))*6^(1/2)-1/2*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2*arctan((3*x^2-1)^(1/4)))*EllipticF(sin(2*arct
an((3*x^2-1)^(1/4))),1/2*2^(1/2))*(1+(3*x^2-1)^(1/2))*(x^2/(1+(3*x^2-1)^(1/2))^2)^(1/2)/x*3^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {443, 325, 234, 220, 400, 442} \[ -\frac {\sqrt [4]{3 x^2-1}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(-2 + 3*x^2)*(-1 + 3*x^2)^(3/4)),x]

[Out]

-(-1 + 3*x^2)^(1/4)/(2*x) + (Sqrt[3/2]*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/4 - (Sqrt[3/2]*ArcTanh[(Sqrt[
3/2]*x)/(-1 + 3*x^2)^(1/4)])/4 - (Sqrt[3]*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[
2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(2*x)

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 234

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/(b*x), Subst[Int[1/Sqrt[1 - x^4/a],
x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 400

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[1/c, Int[1/(a + b*x^2)^(3/4), x],
x] - Dist[d/c, Int[x^2/((a + b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d,
0]

Rule 442

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> -Simp[(b*ArcTan[(Rt[-(b^2/a), 4]*
x)/(Sqrt[2]*(a + b*x^2)^(1/4))])/(Sqrt[2]*a*d*Rt[-(b^2/a), 4]^3), x] + Simp[(b*ArcTanh[(Rt[-(b^2/a), 4]*x)/(Sq
rt[2]*(a + b*x^2)^(1/4))])/(Sqrt[2]*a*d*Rt[-(b^2/a), 4]^3), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0
] && NegQ[b^2/a]

Rule 443

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a]
|| IntegerQ[m/2])

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=\int \left (-\frac {1}{2 x^2 \left (-1+3 x^2\right )^{3/4}}+\frac {3}{2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{x^2 \left (-1+3 x^2\right )^{3/4}} \, dx\right )+\frac {3}{2} \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{-1+3 x^2}}{2 x}-2 \left (\frac {3}{4} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx\right )+\frac {9}{4} \int \frac {x^2}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{-1+3 x^2}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-2 \frac {\left (\sqrt {3} \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{2 x}\\ &=-\frac {\sqrt [4]{-1+3 x^2}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{2 x}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 52, normalized size = 0.35 \[ \frac {\left (1-3 x^2\right )^{3/4} F_1\left (-\frac {1}{2};\frac {3}{4},1;\frac {1}{2};3 x^2,\frac {3 x^2}{2}\right )}{2 x \left (3 x^2-1\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^2*(-2 + 3*x^2)*(-1 + 3*x^2)^(3/4)),x]

[Out]

((1 - 3*x^2)^(3/4)*AppellF1[-1/2, 3/4, 1, 1/2, 3*x^2, (3*x^2)/2])/(2*x*(-1 + 3*x^2)^(3/4))

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fricas [F]  time = 4.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}}{9 \, x^{6} - 9 \, x^{4} + 2 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="fricas")

[Out]

integral((3*x^2 - 1)^(1/4)/(9*x^6 - 9*x^4 + 2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((3*x^2 - 1)^(3/4)*(3*x^2 - 2)*x^2), x)

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maple [F]  time = 2.09, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(3*x^2-2)/(3*x^2-1)^(3/4),x)

[Out]

int(1/x^2/(3*x^2-2)/(3*x^2-1)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(3*x^2-2)/(3*x^2-1)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((3*x^2 - 1)^(3/4)*(3*x^2 - 2)*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,{\left (3\,x^2-1\right )}^{3/4}\,\left (3\,x^2-2\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(3*x^2 - 1)^(3/4)*(3*x^2 - 2)),x)

[Out]

int(1/(x^2*(3*x^2 - 1)^(3/4)*(3*x^2 - 2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(3*x**2-2)/(3*x**2-1)**(3/4),x)

[Out]

Integral(1/(x**2*(3*x**2 - 2)*(3*x**2 - 1)**(3/4)), x)

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